Why this series converges to 0 Let $\alpha$ be an irrational number. Then $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k = 1}^n e^{2 \pi i k\alpha} = 0 = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k = 1}^n e^{-2 \pi i k\alpha}. $$
Please give some ideas of how to verify this !! 
 A: We can show this in an elementary way by using the fact that
$$\sum_{k=1}^{n}z^k = z\sum_{k=0}^{n-1}z^{k} = z\frac{1 - z^n}{1-z}$$
provided that $z \neq 1$. We apply this formula with $z = e^{2\pi i \alpha}$, which is not equal to $1$ since $\alpha$ is irrational. Taking absolute values, we get
$$\begin{aligned}
\left|\frac{1}{n}\sum_{k=1}^{n} e^{2\pi i k \alpha}\right| &= \frac{1}{n}\left|e^{2\pi i \alpha}\frac{1 - e^{2\pi i \alpha n}}{1 - e^{2\pi i \alpha}}\right| \\
&= \frac{1}{n}\left|\frac{\sin(\pi \alpha n)}{\sin(\pi \alpha)}\right| \\
&\leq \frac{1}{n}\frac{1}{|\sin(\pi \alpha)|} \\
\end{aligned}$$
which converges to $0$ as $n \to \infty$. Note that we did not require $\alpha$ to be irrational, only that $e^{2\pi i \alpha} \neq 1$, which is true as long as $\alpha$ is not an integer. (Notice that if $\alpha$ is an integer, the limit is $1$, not $0$.)
A: This follows from equidistributedness of $\alpha,2\alpha,\cdots$. Take $f(x)=e^{2\pi i x}$ on $[0,1]$ then
$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k = 1}^n e^{2 \pi i k\alpha} = \int_0^1 f(x)dx=0$$
See 1 and 2 for instance. 
Note that the second equally follows immediately from the first one.
